What is the integral of sin(3x-1) ?

The integral of sin(3x-1) is -1/3 cos(3x-1)+C

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integral of sin(3x-1)

\int sin (3 x - 1) d x

- \frac{1}{3} cos (3 x - 1) + C

Solution steps

\int sin (3 x - 1) d x

Apply u-substitution

: \int sin (u) \frac{1}{3} d u

= \int sin (u) \frac{1}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{1}{3} (- cos (u))

Substitute back

u = 3 x - 1

= \frac{1}{3} (- cos (3 x - 1))

Simplify

= - \frac{1}{3} cos (3 x - 1)

Add a constant to the solution

= - \frac{1}{3} cos (3 x - 1) + C

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What is the integral of sin(3x-1) ?
The integral of sin(3x-1) is -1/3 cos(3x-1)+C